What Is the Longest Wavelength Enhanced in a Thin Soap Film?

[chipmunk951]

Homework Statement

A soap film is 129 nm thick. The film is in air and illuminated by white light at normal incidence as shown in the figure below. (In the figure the rays have been drawn at an angle to show the multiple reflections and transmissions in the film.) Assume the index of refraction of the soap film is the same as the index of refraction of water n = 1.33. The film is viewed from above and below by a video system which is sensitive to wavelengths from 200 to 1100 nm.
http://img227.imageshack.us/my.php?image=showmeplve2.gif

Homework Equations

1/2 + 2d/lambda = m
1/2 + 2d/lambda = m + 1/2

The Attempt at a Solution

a) When viewed in reflection, what is the longest visible wavelength that will appear enhanced?
b) When viewed in reflection, what is the longest visible wavelength that will have the least intensity?
c) When viewed in transmission, what is the longest visible wavelength that will appear enhanced?

a) Wouldn't this be 1/2 + 2d/lambda = m and solving for lambda where it's inbetween 200-1100 nm?

b) Got this, ended up just being 2nt/m = lambda

c) Not sure on this one, the help provides a link to some slides but the link is broken and I don't think our professor really cares.. Thanks!

 

[anathema]

To answer these questions, we need to use the equation for constructive interference in a thin film: 1/2 + 2d/lambda = m, where d is the thickness of the film, λ is the wavelength of light, and m is an integer representing the order of the interference.

a) When viewed in reflection, the longest visible wavelength that will appear enhanced is when m = 1, giving us λ = 2d. Plugging in the values for d and n, we get λ = 258 nm.

b) When viewed in reflection, the longest visible wavelength that will have the least intensity is when m = 0, giving us λ = 2d. Plugging in the values for d and n, we get λ = 129 nm.

c) When viewed in transmission, the longest visible wavelength that will appear enhanced is when m = 1/2, giving us λ = 4d/3. Plugging in the values for d and n, we get λ = 172 nm. This is because in transmission, the phase change upon reflection at the top and bottom of the film is different, resulting in a different equation for constructive interference (1/2 + 2d/lambda = m + 1/2).

 

[thomate1]

As a scientist, it is important to provide a thorough and accurate response to any question or problem. In this case, the problem is asking about the behavior of a thin soap film under white light illumination. To answer the questions, we need to understand the concept of thin film interference and how it relates to the thickness of the film and the wavelength of light.

First, let's define some important terms. The "thickness" of a thin film is the distance between the top and bottom surfaces of the film. In this case, the soap film is 129 nm thick, meaning it is only 129 billionths of a meter thick. The "index of refraction" is a measure of how much a material bends light as it passes through it. In this problem, we are told that the index of refraction of the soap film is the same as that of water, which is 1.33. Lastly, "white light" is a combination of all visible wavelengths of light, ranging from 400 to 700 nm.

Now, let's look at the three questions.

a) When viewed in reflection, what is the longest visible wavelength that will appear enhanced?

When light hits a thin film, it reflects off the top surface and the bottom surface. Some of the reflected light is in phase (constructive interference) and some is out of phase (destructive interference). The interference pattern created by the reflected light can be seen as different colors or intensities. The longest wavelength that will appear enhanced is the one that experiences constructive interference. In this case, we can use the equation 1/2 + 2d/λ = m, where d is the thickness of the film, λ is the wavelength of light, and m is an integer representing the number of times the light is reflected. We know that d = 129 nm and n = 1.33, so we can solve for λ by setting m = 0 (since we want the first order maximum). This gives us a value of λ = 2d/(n+1) = 2(129 nm)/(1.33+1) = 194.6 nm. Therefore, the longest visible wavelength that will appear enhanced is 194.6 nm.

b) When viewed in reflection, what is the longest visible wavelength that will have the least intensity?

The longest wavelength that will have the least intensity is the one that experiences destructive interference. In this case, we can use the